Find the values of `lambda` ad `mu` if the points `A(-1,4,-2),B(lambda,mu,1)` and `C(0,2,-1)` are collinear.

Answer

To find the values of \( \lambda \) and \( \mu \) such that the points \( A(-1, 4, -2) \), \( B(\lambda, \mu, 1) \), and \( C(0, 2, -1) \) are collinear, we can follow these steps:

### Step 1: Define the vectors
First, we need to find the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \).

The vector \( \overrightarrow{AB} \) is given by:
\[ \overrightarrow{AB} = B - A = (\lambda, \mu, 1) - (-1, 4, -2) = (\lambda + 1, \mu - 4, 3) \]

The vector \( \overrightarrow{AC} \) is given by:
\[ \overrightarrow{AC} = C - A = (0, 2, -1) - (-1, 4, -2) = (1, -2, 1) \]

### Step 2: Set up the condition for collinearity
For points \( A, B, \) and \( C \) to be collinear, the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) must be parallel. This can be expressed using the cross product:
\[ \overrightarrow{AB} \times \overrightarrow{AC} = \mathbf{0} \]

### Step 3: Calculate the cross product
We can calculate the cross product using the determinant of a matrix formed by the components of the vectors:
\[ \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \lambda + 1 & \mu - 4 & 3 \\ 1 & -2 & 1 \end{vmatrix} \]

### Step 4: Expand the determinant
Calculating the determinant, we have:
\[ \overrightarrow{AB} \times \overrightarrow{AC} = \mathbf{i} \left( (\mu - 4) \cdot 1 - 3 \cdot (-2) \right) - \mathbf{j} \left( (\lambda + 1) \cdot 1 - 3 \cdot 1 \right) + \mathbf{k} \left( (\lambda + 1)(-2) - (\mu - 4) \cdot 1 \right) \]

This simplifies to:
\[ \overrightarrow{AB} \times \overrightarrow{AC} = \mathbf{i} \left( \mu - 4 + 6 \right) - \mathbf{j} \left( \lambda + 1 - 3 \right) + \mathbf{k} \left( -2(\lambda + 1) - (\mu - 4) \right) \]
\[ = \mathbf{i} (\mu + 2) - \mathbf{j} (\lambda - 2) + \mathbf{k} (-2\lambda - \mu + 2) \]

### Step 5: Set the components to zero
For the cross product to be the zero vector, each component must equal zero:
1. \( \mu + 2 = 0 \)
2. \( \lambda - 2 = 0 \)
3. \( -2\lambda - \mu + 2 = 0 \)

### Step 6: Solve the equations
From the first equation:
\[ \mu + 2 = 0 \implies \mu = -2 \]

From the second equation:
\[ \lambda - 2 = 0 \implies \lambda = 2 \]

Now, substituting \( \lambda = 2 \) and \( \mu = -2 \) into the third equation to verify:
\[ -2(2) - (-2) + 2 = -4 + 2 + 2 = 0 \]
This holds true.

### Final Answer
Thus, the values of \( \lambda \) and \( \mu \) are:
\[ \lambda = 2, \quad \mu = -2 \]